Simplify and expand the following expression: $ \dfrac{2n - 8}{3n + 8}-\dfrac{3n - 3}{3n + 6} $
Explanation: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3n + 8)(3n + 6)$ Multiply the first term by $\dfrac{3n + 6}{3n + 6}$ $ \begin{align*} \dfrac{2n - 8}{3n + 8} \times \dfrac{3n + 6}{3n + 6} & = \dfrac{(2n - 8)(3n + 6)}{(3n + 8)(3n + 6)} \\ & = \dfrac{6n^2 - 12n - 48}{(3n + 8)(3n + 6)}\end{align*} $ Multiply the second term by $\dfrac{3n + 8}{3n + 8}$ $ \begin{align*} \dfrac{3n - 3}{3n + 6} \times \dfrac{3n + 8}{3n + 8} & = \dfrac{(3n - 3)(3n + 8)}{(3n + 6)(3n + 8)} \\ & = \dfrac{9n^2 + 15n - 24}{(3n + 6)(3n + 8)}\end{align*} $ Now we have: $ = \dfrac{6n^2 - 12n - 48}{(3n + 8)(3n + 6)} - \dfrac{9n^2 + 15n - 24}{(3n + 6)(3n + 8)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{6n^2 - 12n - 48 - (9n^2 + 15n - 24)}{(3n + 8)(3n + 6)} $ $ = \dfrac{6n^2 - 12n - 48 - 9n^2 - 15n + 24}{(3n + 8)(3n + 6)} $ $ = \dfrac{-3n^2 - 27n - 24}{(3n + 8)(3n + 6)}$ Expand the denominator: $ = \dfrac{-3n^2 - 27n - 24}{9n^2 + 42n + 48}$ Simplify: $ = \dfrac{-n^2 - 9n - 8}{3n^2 + 14n + 16}$